Two conducting balls hanging from the same point

1. The problem statement, all variables and given/known data
In the figure, two conducting balls of identical mass m = 20 g and identical charge q hang from nonconducting threads of length L = 100 cm. If x = 5.9 cm, what is q? Since x is much smaller than L approximate sin(θ) by θ.

This is what a classmate of mine posted as a way to solve the problem, but it hasn't been working for me. Does this look like the right method? What does the T before cos and sin mean?
Is it necessary to calculate theta, by sin(theta)=(0.059m/2)/(1 m) ....theta=1.69 deg ?

Three forces work on each of the balls. If you draw them for one of the two you see what your classmate is doing.
You recognize mg and I suppose you also recognize ##kq^2\over x^2##. The balss can only hang still if the threads compensate both of these at the same time. T stands for tension: a thread pulls in a direction along the thread. That's all it can do! Grab one and play with it to confirm.

By the way, in math and in physics angles are measured in radians. For ##0<\alpha<\pi/2## you have ##\sin\alpha < \alpha<\tan\alpha## and for small ##\alpha## one can show ##\sin\alpha \approx\alpha\approx\tan\alpha## (try it on a calulator with radians)

I'm confused. we say tan theta = x/2L, or opposite/adjacent = x/2L. It makes sense that the opposite side is x/2, but that would leave the adjacent side as L, but we've already said the hypotenuse was L.

Hello Iason,
Pretty old thread you are now resuming! Why not start a new one ?
To answer your question: You are not messing up at all, but if you are entitled to use ##\theta## for ##\sin\theta##, then you are also entitled to use ##\theta## for ##\tan\theta## (##\cos\theta=1##).

So: what you say is correct, but for small angles it's OK. (You could check -- not by solving the trigonometric equation, but by comparing ##\arctan(0.059)## with ##\arcsin(0.059)##.

Hello Iason,
Pretty old thread you are now resuming! Why not start a new one ?
To answer your question: You are not messing up at all, but if you are entitled to use ##\theta## for ##\sin\theta##, then you are also entitled to use ##\theta## for ##\tan\theta## (##\cos\theta=1##).

So: what you say is correct, but for small angles it's OK. (You could check -- not by solving the trigonometric equation, but by comparing ##\arctan(0.059)## with ##\arcsin(0.059)##.

And compare the situation if ##\theta = {\pi\over 4}##

Thank BvU!
Sorry, I'm new here and joined because I had a problem similar to this (different given conditions) and stumbled across this while trying to solve it. Thank you! That makes sense.
I don't think I can use that in my situation, but it was good to see!